QRZ! Ham Radio 11/QRZ Ham Radio Callsign Database - Volume 11.iso/files/library/db.txt

Text File | 1998-02-02 | 10.2 KB | 258 lines

>Newsgroups: rec.radio.amateur.misc
>Path: gonix!uunet!newsgate.melpar.esys.com!melpar!phb
>From: [e-mail] (Paul H. Bock)
>Subject: TUTORIAL:  dB & dBm
>Sender: [e-mail] (Melpar News Administrator)
>Organization: E-Systems, Melpar Division
>Date: Tue, 11 Oct 1994 17:23:05 GMT
>Message-ID: <phb.781896185@melpar>
>Lines: 247


                     USING AND UNDERSTANDING DECIBELS

                                    by

                          Paul H. Bock, Jr. K4MSG

Author's Note:  This tutorial was originally written for the use of
non-RF/analog engineers (digital, software) and non-engineers who
needed an easy-to-follow reference on the general use of the decibel.
I hope that some amateur operators may find it useful as well.

     While the historical accuracy of the comments relating to the
telephone company and telephone company engineers may be open to
question (the information as supplied to me was anecdotal), the
technical points made should be valid regardless of the exact turn
of history.


*General*

     The decibel, or dB, is a means of expressing either the gain
of an active device (such as an amplifier) or the loss in a passive
device (such as an attenuator or length of cable).  The decibel was
developed by the telephone company to conveniently express the gain
or loss in telephone transmission systems.  The decibel is best
understood by first discussing the rationale for its development.

     If we have two cascaded amplifiers as shown below, with power
gain factors A1 and A2 as indicted, the total gain is the product
of the individual gains, or A1 x A2.

     Input >-------- Amp #1 --------- Amp #2 ------> Output

                    A1 = 275          A2 = 55

     In the example, the total gain factor At = 275 x 55 = 15,125.
Now, imagine for a moment what it would be like to calculate the
total gain of a string of amplifiers.  It would be a cumbersome
task at best, and especially so if there were portions of the
cascade which were lossy and reduced the total gain, thereby
requiring division as well as multiplication.

     It was for the reason stated above that Bell Telephone
developed the decibel.  Thinking back to the rules for logarithms,
we recall that rather than multiplying two numbers we can add their
logarithms and then take the antilogarithm of this sum to find the
product we would have gotten had we multiplied the two numbers.
Mathematically,
                     log (A x B) = log A + log B

     If we want to divide one number into another, we subtract the
logarithm of the divisor from the logarithm of the dividend, or in
other words
                     log (A/B) = log A - log B

     The telephone company decided that it might be convenient to
handle gains and losses this way, so they invented a unit of gain
measurement called a "Bel," named after Alexander Graham Bell.
They defined the Bel as

                     Gain in Bels = log A

           where  A = Power amplification factor

     Going back to our example, we find that log 275 = 2.439 and
log 55 = 1.740, so the total gain in our cascade is

                     2.439 + 1.74 = 4.179 Bels

     It quickly occurred to the telephone company engineers that
using Bels meant they would be working to at least two decimal
places.  They couldn't just round things off to one decimal place,
since 4.179 bels is a power gain of 15,101 while 4.2 bels is a
power gain of 15,849, yielding an error of about 5%.  At that point
it was decided to express power gain in units which were equal to
one-tenth of a Bel, or in deci-Bels.  This simply meant that the
gain in Bels would be multiplied by 10, since there would be ten
times more decibels than Bels.  This changes the formula to

            Gain in decibels (dB) = 10 log A    (Eq. 1)

     Again using our example, the gain in the cascade is now

                    24.39 + 17.40 = 41.79 decibels

     The answer above is accurate, convenient to work with, and can
be rounded off to the first decimal place will little loss in
accuracy; 41.79 dB is a power gain of 15,101, while 41.8 dB is a
power gain of 15,136, so the error is only 0.23%.

     What if the power gain factor is less than one, indicating an
actual power loss?  The calculation is performed as shown above
using Equation 1, but the result will be different.  Suppose we
have a device whose power gain factor is 0.25, which means that it
only outputs one-fourth of the power fed into it?  Using Equation
1, we find
                     G = 10 log (0.25)

                     G = 10 (-0.60)

                     G = -6.0 dB

     The minus sign occurs because the logarithm of any number less
than 1 is always negative.  This is convenient, since a power loss
expressed in dB will always be negative.

     There are two common methods of using the decibel.  The first
is to express a known power gain factor in dB, as just described.
The second is to determine the power gain factor and convert it to
dB, which can all be done in one calculation.  The formula for this
operation is as follows:
                                  Po
                      G = 10 log ----          (Eq. 2)
                                  Pi

               where G  = Gain in dB
                     Po = Power output from the device
                     Pi = Power input to the device

     Both Po and Pi should be in the same units; i.e., watts,
milliwatts, etc.  Note that Equation 2 deals with power, not
voltage or current; these are handled differently when converted
to dB, and are not relevant to this discussion.  Below are two
examples of the correct application of Equation 2:

     Ex. 1:  An amplifier supplies 3.5 watts of output with an
           input of 20 milliwatts.  What is the gain in dB?

                                3.5 watts
                    G = 10 log ----
                                0.02 watts

                    G = 10 log (175)

                    G = 10 (2.24)

                    G = 22.4 dB


     Ex. 2:  A length of coaxial transmission line is being fed
           with 150 watts from a transmitter, but the power
           measured at the output end of the line is only 112
           watts.  What is the line loss in dB?

                               112 watts
                    G = 10 log ---
                               150 watts

                    G = 10 log 0.747

                    G = 10 (-0.127)

                    G = -1.27 dB


*Non-relative (Absolute) Uses of the Decibel*

     The most common non-relative, or absolute, use of the decibel
is the dBm, or decibel relative to one milliwatt.  It is different
from the dB because it represents, in physical terms, an absolute
amount of power which can be measured.

     The difference between "relative" and "absolute" can be
understood easily by considering temperature.  For example, if I
say that it is "20 degrees colder now than it was this morning,"
it's a relative measurement; unless the listener knows how cold it
was this morning, it doesn't mean anything in absolute terms.  If,
however, I say, "It was 20 degrees C this morning, but it's 20
degrees colder now," then the listener knows exactly what is meant;
it is now 0 degrees C.  This can be measured on a thermometer and
is referenced to an absolute temperature scale.

     So it is with dB and dBm.  A dB is merely a relative
measurement, while a dBm is referenced to an absolute quantity:
the milliwatt (1/1000 of a watt).  We can apply this concept to
Equation 1 as follows:

                     dBm = 10 log (P) (1000 mW/watt)

               where dBm = Power in dB referenced to 1 milliwatt
                       P = Power in watts

     For example, take the case where we have a power level of 1
milliwatt:

                 dBm = 10 log (0.001 watt) (1000 mW/watt)

                 dBm = 10 log (1)

                 dBm = 10 (0)

                 dBm = 0

     Thus, we see that a power level of 1 milliwatt is 0 dBm.  This
makes sense intuitively, since our reference power level is also
1 milliwatt.  If the power level was 1 watt, however, we find that

                 dBm = 10 log (1 watt) (1000 mW/watt)

                 dBm = 10 (3)

                 dBm = 30

     The dBm can also be negative, just like the dB; if our power
level is 1 microwatt, we find that

                 dBm = 10 log (1 x 10E-6 watt) (1000 mW/watt)

                 dBm = -30 dBm

     Since the dBm is an absolute amount of power, it can be
converted back to watts if desired.   Since it is in logarithmic
form it may also be conveniently combined with other dB terms,
making system analysis easier.  For example, suppose we have a
signal source with an output power of -70 dBm, which we wish to
connect to an amplifier having 22 dB gain through a cable having
8.5 dB loss.  What is the output level from the amplifier?  To find
the answer, we just add the gains and losses as follows:

                 Output = -70 dBm + 22 dB + (-8.5 dB)

                 Output = -70 dBm + 22 dB - 8.5 dB

                 Output = -56.5 dBm

     As a final note, power level may be referenced to other
quantities and expressed in dB form.  Below are some examples:

         dBW = Power level referenced to 1 watt

         dBk = Power level referenced to 1 kilowatt (1000 watts)

     One other common usage is dBc, which is a relative term like
dB alone.  It means "dB referenced to a carrier level" and is most
commonly seen in receiver specifications regarding spurious signals
or images.  For example, "Spurious signals shall not exceed -50
dBc" means that spurious signals will always be at least 50 dB less
than some specified carrier level present (which could mean "50 dB
less than the desired signal").

       * Paul H. Bock, Jr. K4MSG  * Principal Systems Engineer
(|_|)  * E-Systems/Melpar Div.    * Telephone: [phone] x2062
 | |)  * 7700 Arlington Blvd.     * Internet: [e-mail]
       * Falls Church, VA 22046   * Mailstop: N203

 "Imagination is more important than knowledge." - Albert Einstein